QJEEM24Ma5fd1: JEE Main 2024 Mathematics - Class 12 JEEM24Ma5fd1 Solution: If the system of equations has infinitely many solutions, then…

$\begin{aligned}& x+4 y-z=\lambda \\& 7 x+9 y+\mu z=-3 \\& 5 x+y+2 z=-1 \\& {\left[\begin{array}{ccc}1 & 4 & -1 \\7 & 9 & \mu \\5 & 1 & 2\end{array}\right]\left[\begin{array}{l}x \\y \\z\end{array}\right]=\left[\begin{array}{c}\lambda \\-3 \\-1\end{array}\right]} \\& A=\left[\begin{array}{lll}1 & 4 & -1 \\7 & 9 & \mu \\5 & 1 & 2\end{array}\right], B=\left[\begin{array}{c}\lambda \\-3 \\-1\end{array}\right] \\& A X=B \\& X=A^{-1} B \\& \quad=\frac{\operatorname{adj} A}{|A|} B\end{aligned}$

If $|A|=0$ and $(\operatorname{adj} A) \cdot B=0$ , system has infinitely many solutions.

$\begin{aligned}& |A|=18-\mu-4(14-5 \mu)-1(7-45)=0 \\& \Rightarrow 18-\mu-56+20 \mu+38=0 \\& \Rightarrow 19 \mu=0 \\& \Rightarrow \mu=0 \\& \text { Also adjA }=\left[\begin{array}{ccc}18 & -9 & 9 \\-14 & 7 & -7 \\-38 & 19 & -19\end{array}\right] \\& (\text { adj } A) \cdot B=0 \\& {\left[\begin{array}{ccc}18 & -9 & 9 \\-14 & 7 & -7 \\-38 & 19 & -19\end{array}\right]\left[\begin{array}{c}\lambda \\-3 \\-1\end{array}\right]=\left[\begin{array}{l}0 \\0 \\0\end{array}\right]} \\& 18 \lambda+27-9=0 \\& \Rightarrow 18 \lambda=-18 \\& \Rightarrow \lambda=-1 \\& \Rightarrow 2 \mu+3 \lambda=3(-1)=-3\end{aligned}$